Relational Algebra
Lecture 2
Relational Model
•
•
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•
•
•
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Basic Notions
Fundamental Relational Algebra Operations
Additional Relational Algebra Operations
Extended Relational Algebra Operations
Null Values
Modification of the Database
Views
Bags and Bag operations
Basic Structure
• Formally, given sets D1, D2, …. Dn a relation r is a subset of
D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples (a1, a2, …, an) where each ai  Di
• Example:
customer_name = {Jones, Smith, Curry, Lindsay}
customer_street = {Main, North, Park}
customer_city
= {Harrison, Rye, Pittsfield}
Then r = { (Jones, Main, Harrison),
(Smith, North, Rye),
(Curry, North, Rye),
(Lindsay, Park, Pittsfield) }
is a relation over
customer_name , customer_street, customer_city
Attribute Types
• Each attribute of a relation has a name
• The set of allowed values for each attribute is called the
domain of the attribute
• Attribute values are (normally) required to be atomic; that
is, indivisible
– Note: multivalued attribute values are not atomic
({secretary. clerk}) is example of multivalued attribute
position
– Note: composite attribute values are not atomic
• The special value null is a member of every domain
• The null value causes complications in the definition of
many operations
– We shall ignore the effect of null values in our main
presentation and consider their effect later
Relation Schema
• A1, A2, …, An are attributes
• R = (A1, A2, …, An ) is a relation schema
Example:
Customer_schema = (customer_name, customer_street,
customer_city)
• r(R) is a relation on the relation schema R
Example:
customer (Customer_schema)
Relation Instance
• The current values (relation instance) of a relation are
specified by a table
• An element t of r is a tuple, represented by a row in a
attributes
table
customer_name customer_street
customer_city
Jones
Smith
Curry
Lindsay
Harrison
Rye
Rye
Pittsfield
Main
North
North
Park
customer
(or columns)
tuples
(or rows)
Database
• A database consists of multiple relations
• Information about an enterprise is broken up into parts,
with each relation storing one part of the information
account : stores information about accounts
depositor : stores information about which customer
owns which account
customer : stores information about customers
• Storing all information as a single relation such as
bank(account_number, balance, customer_name, ..)
results in repetition of information (e.g., two customers
own an account) and the need for null values (e.g.,
represent a customer without an account)
Query Languages
• Language in which user requests information from the
database.
• Categories of languages
– Procedural
– Non-procedural, or declarative
• “Pure” Procedural languages:
– Relational algebra
– Tuple relational calculus
– Domain relational calculus
• Pure languages form underlying basis of query languages
that people use.
What is “algebra”
• Mathematical model consisting of:
– Operands --- Variables or values;
– Operators --- Symbols denoting procedures that
construct new values from a given values
• Relational Algebra is algebra whose operands are
relations and operators are designed to do the most
commons things that we need to do with relations
Basic Relational Algebra Operations
•
•
•
•
•
Select
Project
Union
Set Difference (or Substract or minus)
Cartesian Product
Select Operation
• Notation:  p(r)
• p is called the selection predicate
• Defined as:
p(r) = {t | t  r and p(t)}
Where p is a formula in propositional calculus
consisting of terms connected by :  (and),  (or),
 (not)
Each term is one of:
<attribute>op <attribute> or <constant>
where op is one of: =, , >, . <. 
• Example of selection:
Account(account_number, branch_name,balance)
 branch-name=“Perryridge”(account)
Select Operation – Example
• Relation r
• A=B ^ D > 5 (r)
A
B
C
D


1
7


5
7


12
3


23 10
A
B
C
D


1
7


23 10
Project Operation
• Notation:
A1, A2, …, Ak (r)
where A1, A2 are attribute names and r is a relation.
• The result is defined as the relation of k columns obtained
by erasing the columns that are not listed
• Duplicate rows removed from result, since relations are sets
• E.g. to eliminate the branch-name attribute of account
account-number, balance (account)
• If relation Account contains 50 tuples, how many tuples contains
account-number, balance (account) ?
• If relation Account contains 50 tuples, how many tuples contains
, balance (account) ?
Project Operation – Example
• Relation r:
• A,C (r)
A
A
B
C

10
1

20
1

30
1

40
2
C

1

1

1

2
A
=
C

1

1

2
That is, the projection of
a relation on a set of
attributes is a set of tuples
Union Operation
• Consider relational schemas:
Depositor(customer_name, account_number)
Borrower(customer_name, loan_number)
• For r  s to be valid.
1. r, s must have the same number of attributes
2. The attribute domains must be compatible (e.g., 2nd
column of r deals with the same type of values as does the
2nd column of s)
Find all customers with either an account or a loan
customer-name (depositor)  customer-name (borrower)
Union Operation
• Notation: r  s
• Defined as:
r  s = {t | t  r or t  s}
Union Operation – Example
• Relations r, s:
A
B
A
B

1

2

2

3

1
s
r
r  s:
A
B

1

2

1

3
Set Difference Operation
• Notation r – s
• Defined as:
r – s = {t | t  r and t  s}
• Set differences must be taken between compatible
relations.
– r and s must have the same number of attributes
– attribute domains of r and s must be compatible
Set Difference Operation – Example
• Relations r, s:
A
B
A
B

1

2

2

3

1
s
r
r – s:
A
B

1

1
Cartesian-Product Operation
• Notation r x s
• Defined as:
r x s = {t q | t  r and q  s}
• Assume that attributes of r(R) and s(S) are disjoint. (That
is, R  S =  ).
• If attributes of r(R) and s(S) are not disjoint, then renaming
must be used.
Cartesian-Product Operation-Example
Relations r, s:
A
B
C
D
E

1

2




10
10
20
10
a
a
b
b
r
s
r x s:
A
B
C
D
E








1
1
1
1
2
2
2
2








10
10
20
10
10
10
20
10
a
a
b
b
a
a
b
b
Composition of Operations
• Can build expressions using multiple operations
• Example: A=C(r  s)
• rs
• A=C(r  s)
A
B
C
D
E








1
1
1
1
2
2
2
2








10
10
20
10
10
10
20
10
a
a
b
b
a
a
b
b
A
B
C
D
E



1
2
2
 10
 20
 20
a
a
b
Rename Operation
• Allows us to name, and therefore to refer to, the results of
relational-algebra expressions.
• Allows us to refer to a relation by more than one name.
Example:
 X (E)
returns the expression E under the name X
If a relational-algebra expression E has arity n, then
 (A1, A2, …, An) (E)
xx E under the name X, and with
returns the result of expression
the attributes renamed to A1, A2, …., An.
Banking Example
branch (branch-name, branch-city, assets)
customer (customer-name, customer-street, customercity)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)
Keys
• Let K  R
• K is a superkey of R if values for K are sufficient to identify
a unique tuple of each possible relation r(R)
– by “possible r ” we mean a relation r that could exist in
the enterprise we are modeling.
– Example: {customer_name, customer_street} and
{customer_name}
are both superkeys of Customer, if no two customers can
possibly have the same name.
• K is a candidate key if K is minimal
Example: {customer_name} is a candidate key for.
• Primary Key
Keys
Superkeys
Candidate
keys K
Primary key
Example Queries
•
Find all loans of over $1200
amount > 1200 (loan)
•
Find the loan number for each loan of an amount greater than
$1200
loan-number (amount > 1200 (loan))
Example Queries
• Find the names of all customers who have a loan, an
account, or both, from the bank
customer-name (borrower)  customer-name (depositor)
• Find the names of all customers who have a loan and an
account at bank.
customer-name (borrower)  customer-name (depositor)
Additional Operations
We define additional operations that do not add any power
to the relational algebra, but that simplify common queries.
• Set intersection
• Natural join
• Division
• Assignment
Set-Intersection Operation
Notation: r  s
Defined as:
r  s ={ t | t  r and t  s }
Assume:
– r, s have the same arity
– attributes of r and s are compatible
• Note: r  s = r - (r - s)
•
•
•
•
Set-Intersection Operation - Example
• Relation r, s:
A



B
1
2
1
r
• rs
A
B


2
3
s
A
B

2
Natural-Join Operation
Notation: r s
• Let r and s be relations on schemas R and S respectively.
Then, r s is a relation on schema R  S obtained as follows:
– Consider each pair of tuples tr from r and ts from s.
– If tr and ts have the same value on each of the attributes in R
 S, add a tuple t to the result, where
• t has the same value as tr on r
• t has the same value as ts on s
• Example:
R = (A, B, C, D)
S = (E, B, D)
– Result schema = (A, B, C, D, E)
– r s is defined as:
r.A, r.B, r.C, r.D, s.E (r.B = s.B  r.D = s.D (r x s))

Natural Join Operation – Example
• Relations r, s:
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
r
s
s
A
B
C
D
E





1
1
1
1
2





a
a
a
a
b





Natural Join Example
sid
bid
day
22
58
101
103
10/ 1 0/ 9 6
11/ 1 2/ 9 6
sid
sn am e
rat ing
ag e
22
dus tin
7
45.0
31
lub b er
8
55.5
58
rusty
10
35.0
R1
R1
S1
S1 =
sid
sname
rating
age
bid
day
22
58
dustin
rusty
7
10
45.0
35.0
101
103
10/10/96
11/12/96
Other Types of Joins
• Condition Join (or “theta-join”):
R  c S   c ( R  S )
• Result schema same as that of crossproduct.
• May have fewer tuples than cross-product.
• Equi-Join: Special case: condition c
contains only conjunction of equalities.

“Theta” Join Example
sid
bid
day
sid
sname
rating
age
22
58
101
103
10/10/96
11/12/96
22
dustin
7
45.0
31
58
lubber
rusty
8
10
55.5
35.0
R1
S1
S1
S1.sid  R1.sid
R1
=
(sid)
sname
rating
age
(sid)
bid
day
22
31
dustin
lubber
7
8
45.0
55.5
58
58
103
103
11/12/96
11/12/96
Division Operation
Notation:
rs
• Suited to queries that include the phrase “for all”.
• Let r and s be relations on schemas R and S
respectively where
– R = (A1, …, Am, B1, …, Bn)
– S = (B1, …, Bn)
The result of r  s is a relation on schema
R – S = (A1, …, Am)
r  s = { t | t   R-S(r)   u  s ( tu  r ) }
Division Operation – Example
Relations r, s:
r  s:
A


A
B
B











1
2
3
1
1
1
3
4
6
1
2
1
r
2
s
Another Division Example
Relations r, s:
A
B
C
D
E
D
E








a
a
a
a
a
a
a
a








a
a
b
a
b
a
b
b
1
1
1
1
3
1
1
1
a
b
1
1
r
r  s:
A
B
C


a
a


s
Division Operation
• Definition in terms of the basic algebra operation
Let r(R) and s(S) be relations, and let S  R
r  s = R-S (r) –R-S ( (R-S (r) x s) – R-S,S(r))
To see why
– R-S,S(r) simply reorders attributes of r
– R-S(R-S (r) x s) – R-S,S(r)) gives those tuples t in
R-S (r) such that for some tuple u  s, tu  r.
Assignment Operation
• The assignment operation () provides a convenient
way to express complex queries.
– Write query as a sequential program consisting of
• a series of assignments
• followed by an expression whose value is
displayed as a result of the query.
– Assignment must always be made to a temporary
relation variable.
• Example: Write r  s as
temp1  R-S (r)
temp2  R-S ((temp1 x s) – R-S,S (r))
result = temp1 – temp2
Extended Relational-Algebra-Operations
• Generalized Projection
• Outer Join
• Aggregate Functions
Generalized Projection
• Extends the projection operation by allowing arithmetic
functions to be used in the projection list.
 F1, F2, …, Fn(E)
• E is any relational-algebra expression
• Each of F1, F2, …, Fn are are arithmetic expressions
involving constants and attributes in the schema of E.
• Given relation credit-info(customer-name, limit, creditbalance), find how much more each person can spend:
customer-name, limit – credit-balance (credit-info)
Aggregate Functions and Operations
• Aggregation function takes a collection of values and
returns a single value as a result.
avg: average value
min: minimum value
max: maximum value
sum: sum of values
count: number of values
• Aggregate operation in relational algebra
G1, G2, …, Gn g F1( A1), F2( A2),…, Fn( An) (E)
– E is any relational-algebra expression
– G1, G2 …, Gn is a list of attributes on which to group
(can be empty)
– Each Fi is an aggregate function
– Each Ai is an attribute name
Aggregate Operation – Example
• Relation
r:
g sum(c) (r)
A
B
C








7
sum-C
27
7
3
10
Aggregate Operation – Example
• Relation account grouped by branch-name:
branch-name account-number
Perryridge
Perryridge
Brighton
Brighton
Redwood
branch-name
g
balance
A-102
A-201
A-217
A-215
A-222
sum(balance)
400
900
750
750
700
(account)
branch-name
Perryridge
Brighton
Redwood
balance
1300
1500
700
Aggregate Functions
• Result of aggregation does not have a name
– Can use rename operation to give it a name
– For convenience, we permit renaming as part of
aggregate operation
branch-name
g sum(balance) as sum-balance (account)
Outer Join – Example
• Relation loan
loan-number
branch-name
L-170
L-230
L-260
Downtown
Redwood
Perryridge
amount
3000
4000
1700
 Relation borrower
customer-name loan-number
Jones
Smith
Hayes
L-170
L-230
L-155
Outer Join
• An extension of the join operation that avoids loss of
information.
• Computes the join and then adds tuples form one relation
that does not match tuples in the other relation to the result
of the join.
• Uses null values:
– null signifies that the value is unknown or does not exist
– All comparisons involving null are (roughly speaking)
false by definition.
• We shall study precise meaning of comparisons with
nulls later
Left Outer Join
• Join
loan
Borrower
loan-number
L-170
L-230
branch-name
Downtown
Redwood
amount
customer-name
3000
4000
Jones
Smith
amount
customer-name
 Left Outer Join
loan
Borrower
loan-number
L-170
L-230
L-260
branch-name
Downtown
Redwood
Perryridge
3000
4000
1700
Jones
Smith
null
Right Outer Join, Full Outer Join
• Right Outer Join
loan
borrower
loan-number
L-170
L-230
L-155
branch-name
Downtown
Redwood
null
amount
3000
4000
null
customer-name
Jones
Smith
Hayes
Outer Join
loan
borrower
loan-number
L-170
L-230
L-260
L-155
branch-name
Downtown
Redwood
Perryridge
null
amount
3000
4000
1700
null
customer-name
Jones
Smith
null
Hayes
Null Values
• It is possible for tuples to have a null value, denoted by
null, for some of their attributes
• null signifies an unknown value or that a value does not
exist.
• The result of any arithmetic expression involving null is
null.
• Aggregate functions simply ignore null values
• For duplicate elimination and grouping, null is treated like
any other value, and two nulls are assumed to be the same
Null Values
• Comparisons with null values return the special truth value
unknown
– If false was used instead of unknown, then not (A < 5)
would not be equivalent to A >= 5
• Three-valued logic using the truth value unknown:
– OR: (unknown or true)
= true,
(unknown or false)
= unknown
(unknown or unknown) = unknown
– AND: (true and unknown)
= unknown,
(false and unknown)
= false,
(unknown and unknown) = unknown
– NOT: (not unknown) = unknown
• Result of select predicate is treated as false if it evaluates
to unknown
Modification of the Database
• The content of the database may be
modified using the following operations:
– Deletion
– Insertion
– Updating
• All these operations are expressed using
the assignment operator.
Deletion
• A delete request is expressed similarly to a query,
except instead of displaying tuples to the user, the
selected tuples are removed from the database.
• Can delete only whole tuples; cannot delete values on
only particular attributes
• A deletion is expressed in relational algebra by:
rr–E
where r is a relation and E is a relational algebra
query.
Deletion Examples
• Delete all account records in the Perryridge branch.
account  account – branch_name = “Perryridge” (account )
•
Delete all loan records with amount in the range of 0 to 50
loan  loan –  amount 0 and amount  50 (loan)

Delete all accounts at branches located in Needham.
r1   branch_city = “Needham” (account
branch )
r2  branch_name, account_number, balance (r1)
r3   customer_name, account_number (r2
account  account – r2
depositor  depositor – r3
depositor)
Insertion
• To insert data into a relation, we either:
– specify a tuple to be inserted
– write a query whose result is a set of tuples to be
inserted
• in relational algebra, an insertion is expressed by:
r r  E
where r is a relation and E is a relational algebra
expression.
• The insertion of a single tuple is expressed by letting E be
a constant relation containing one tuple.
Insertion Examples
• Insert information in the database specifying that Smith
has $1200 in account A-973 at the Perryridge branch.
account  account  {(“Perryridge”, A-973, 1200)}
depositor  depositor  {(“Smith”, A-973)}

Provide as a gift for all loan customers in the Perryridge
branch, a $200 savings account. Let the loan number serve
as the account number for the new savings account.
r1  (branch_name = “Perryridge” (borrower
loan))
account  account  branch_name, loan_number,200 (r1)
depositor  depositor  customer_name, loan_number (r1)
Updating
• A mechanism to change a value in a tuple without changing
all values in the tuple
• Use the generalized projection operator to do this task
r   F ,F
1
2
, ,F l ,
(r )
• Each Fi is either
– the i th attribute of r, if the ith attribute is not updated, or,
– if the attribute is to be updated Fi is an expression,
involving only constants and the attributes of r, which
gives the new value for the attribute
Update Examples
• Make interest payments by increasing all balances by 5
percent.
account   account_number, branch_name, balance * 1.05 (account)
 Pay all accounts with balances over $10,000 6 percent interest
and pay all others 5 percent
account   account_number, branch_name, balance * 1.06 ( BAL  10000 (account ))
  account_number, branch_name, balance * 1.05 (BAL  10000
(account))
Example Queries
• Find the names of all customers who have a loan at the
Perryridge branch.
customer-name (branch-name=“Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan)))
• Find the names of all customers who have a loan at the
Perryridge branch but do not have an account at any branch of the
bank.
customer-name (branch-name = “Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan))) –
customer-name(depositor)
Example Queries
• Find the largest account balance
1. Rename account relation as d
2. The query is:
balance(account) - account.balance
(account.balance < d.balance (account x d (account)))
Example Queries
• Find all customers who have an account from the
“Downtown” and the Uptown” branches.
Query 1
CN(BN=“Downtown”(depositor
account)) 
CN(BN=“Uptown”(depositor
account))
where CN denotes customer-name and BN denotes
branch-name.
Example Queries
• Find all customers who have an account at all branches
located in Brooklyn city.
customer-name, branch-name (depositor account)
 branch-name (branch-city = “Brooklyn” (branch))
Exercises
•
•
•
•
Employee(ename,str,city)
Works(ename,cname,sal)
Company(cname,city)
Manages(ename,mname)
Employee
Joe
Pine
Mike
Pine
Carol
Oak
Kent
Canton
Kent
Matt
Lucy
Sean
Cleveland
Kent
Kent
Main
Pine
Pine
Manages
Joe
Lucy
Mike
Lucy
Carol
Matt
Lucy Matt
Sean
Lucy
Works
Joe GE 30K
Mike GE 100K
Lucy GE 60K
Sean GE 40K
Carol GE 70K
Matt GE 40K
Company
GE Cleveland
IBM NYC
Find names of employees that live in the same city
and the same street as their managers
•
Employee
(Employee
Joe Pine Kent
Mike Pine Canton
Carol Oak Kent
Lucy Pine Kent
Sean Pine Kent
Manages:
Manages)
Lucy
Lucy
Matt
Matt
Lucy
Employee2
Where mname=employee2.ename & street =employee2.street & city=employee2.street
Joe Pine Kent
Mike Pine Canton
Carol Oak Kent
Lucy Pine Kent
Sean Pine Kent
Lucy
Lucy
Matt
Matt
Lucy
Project on ename: Joe
Sean
Pine Kent
Pine Kent
Main Cleveland
Main Cleveland
Pine Kent
Joe Pine Kent Lucy Pine Kent
Sean Pine Kent Lucy Pine Kent
Find Employees that make more than their managers
•
Works
Manages:
(Works
Manages)
Joe GE
Mike GE
Carol GE
Lucy GE
Sean GE
30K
100K
70K
60K
40K
Lucy
Lucy
Matt
Matt
Lucy
Works2
Where mname=works2.ename &salary >works2.salary
Joe
Mike
Carol
Lucy
Sean
GE 30K Lucy GE 60K
GE 100K Lucy GE 60K
GE 70K Matt GE 40K
GE 60K Matt GE 40K
GE 40K Lucy GE 60K
Project on ename: Mike
Carol
Lucy
Mike GE 100K Lucy GE 60K
Carol GE 70K Matt GE 40K
Lucy GE 60K Matt GE 40K
Find all employees who make more money than any other
employee
Project Works on ename: Joe
Lucy
Sean
Carol
Mike
Matt
•
Works
Works2
Where sal<works2.sal
Project on ename: Joe
Lucy
Sean
Carol
Matt
Joe GE
Joe GE
Joe GE
Joe GE
Joe GE
Lucy GE
Lucy GE
Sean GE
Sean GE
Sean GE
Carol GE
Matt GE
Matt GE
Matt GE
30K Mike GE 100K
30K Carol GE 70K
30K Lucy GE 60K
30K Sean GE 40K
30K Matt GE 40K
60K Carol GE 70K
60K Mike GE 100K
40K Mike GE 100K
40K Carol GE 70K
40K Lucy GE 60K
70K Mike GE 100K
40K Mike GE 100K
40K Carol GE 70K
40K Lucy GE 60K
Substract from first projection the second one: Mike
Find all employees that live in the same city as their company
Works
Company
Employee
Cname=company.cname & ename=employee.ename & city=works.city
Matt GE 40K
• Project on ename: Matt
Cleveland Main Cleveland
Extra Material
Expression Trees
Leaves are operands --- either variables standing for relations or
particular relations
Interior nodes are operators applied to their descendents
customer-name, branch-name
depositor
account
Relational Algebra on Bags
• A bag is like a set but it allows elements to
be repeated in a set.
• Example: {1, 2, 1, 3, 2, 5, 2} is a bag.
• Difference between a bag and a list is that
order is not important in a bag.
• Example: {1, 2, 1, 3, 2, 5, 2} and
{1,1,2,3,2,2,5} is the same bag
Need for Bags
• SQL allows relations with repeated tuples. Thus SQL is not
a relational algebra but rather “bag” algebra
• In SQL one need to specifically ask to remove duplicates,
otherwise replicated tuples will not be eliminated
• Operation projection is more efficient on bags than on sets
Operations on Bags
• Select applies to each tuple and no duplicates are
eliminated
• Project also applies to each tuple and duplicates are not
eliminated. Example
A B C
1 2 3
1 2 5
2 3 7
Projection on A, B
A
1
1
2
B
2
2
3
Other Bag Operations
• An element in the union appears the number of times it
appears in both bags
• Example: {1, 2, 3, 1} UNION {1, 1, 2, 3, 4, 1} =
{1, 1, 1, 1, 1, 2, 2, 3, 3, 4}
• An element appears in the intersection of two bags is the
minimum of the number of times it appears in either.
• Example (con’t): {1, 2, 3, 1} INTERSECTION
{1, 1, 2, 3, 4, 1} = {1, 1, 2, 3}
• An element appears in the difference of two bags A and B
as it appears in A minus the number of times it appears in
B but never less that 0 times
Bag Laws
• Not all laws for set operations are valid for bags:
• Commutative law for union does hold for bags:
R UNION S = S UNION R
• However S union S = S for sets and it is not equal to S if S
is a bag
•
Examples
Sailors
Reserves
sid
bid
day
sid
sn am e
rat ing
ag e
22
58
101
103
10/ 1 0/ 9 6
11/ 1 2/ 9 6
22
dus tin
7
45.0
31
lub b er
8
55.5
58
rusty
10
35.0
•
Boats
bid
101
102
103
104
b n am e
Inte rlake
Inte rlake
C lip p er
Ma rin e
co lor
B lue
R ed
Gr een
R ed
Find names of sailors who’ve reserved boat
#103
• Solution 1:
• Solution 2:
 snam e ((
bid  103
R e serves )  Sailors )
 snam e (
( R e serves  Sailors ))
bid  103
Find names of sailors who’ve reserved a
red boat
• Information about boat color only available
in Boats; so need an extra join:
 snam e ((
B oats )  R e serves  Sailors )
color ' red '

A more efficient (???) solution:
 sname (
(( 
(
Boats ))
sid
bid
color 'red '
Re s)
Sailors )

 A query optimizer can find this given the first solution!
Find sailors who’ve reserved a red or a green
boat
• Can identify all red or green boats, then find
sailors who’ve reserved one of these boats:
 (T em pboats, (
color ' red '  color  ' green '
B oats ))
 snam e (T em pboats  R e serves  Sailors )
Find sailors who’ve reserved a red and a
green boat
• Previous approach won’t work! Must identify
sailors who’ve reserved red boats, sailors
who’ve reserved green boats, then find the
intersection (note that sid is a key for Sailors):
 (T em pred , 
sid
 (Tem pgreen, 
( (
sid
color ' red '
((
B oats )  R e serves ))
color ' green '
B oats )  R e serves ))
 snam e ((T em pred  T em pgreen )  Sailors )
Find the names of sailors who’ve reserved all
boats
• Uses division; schemas of the input
relations to / must be carefully chosen:
 (T em psids, ( 
sid , bid
R e serves ) / ( 
bid
B oats ))
 snam e (T em psids  Sailors )

To find sailors who’ve reserved all ‘Interlake’ boats:
.....
/
bid
(
bnam e ' Interlake'
B oats )
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Relational Algebra